# $V$, $W$ are varieties. Does $V\times \mathbf{P}^1=W\times \mathbf{P}^1$ imply $V=W$?

If $\mathbf{P}^1$ is replaced by the affine line $\mathbf{A}^1$, this becomes the cancellation problem, and we have a pair of famous Danielewski surfaces ($xy=1-z^2$ and $x^2y=1-z^2$) as a counterexample (though I'm still seeking how to prove that..). I suppose in my case this counterexample might no longer work.

Also one may replace $\mathbf{P}^1$ by $\mathbf{P}^n$ or other fixed varieties, or ask again after imposing some conditions on $V$ and $W$ (for example dimensions) if there are counterexamples for my question. And more wildly I may ask for what kind of family $X_n$, we will have the result that $V\times X_n=W\times X_n$ implies $V=W$. Any result of these kind of variations of the problem is also welcomed.

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This problem was studied by Fujita in his paper "Cancellation problem of complete varieties", Inventiones Mathematicae 64 (1981).

He showed that the obstruction to cancellation is caused by the Picard schemes, proving the following remarkable result (see Corollary 7 in the cited paper):

Let $M$, $V$ and $W$ be compact complex manifolds such that $M \times V \cong M \times W$. Assume that $M$ is projective and that $\textrm{Alb}(M)=0$ or $\textrm{Alb}(V)=0$. Then $V \cong W$.

In particular, cancellation problem has a positive answer for $M=\mathbf{P}^n$.

The condition on the Albanese variety is a necessary one; in fact, it is known that cancellation is not always true for abelian varieties.

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WOW, that's amazing! –  lethe Oct 15 '11 at 11:51